You have found the following ages (in years) of all 4 meerkats at your local zoo: $ 11,\enspace 1,\enspace 2,\enspace 2$ What is the average age of the meerkats at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 4 meerkats at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{11 + 1 + 2 + 2}{{4}} = {4\text{ years old}} $ Find the squared deviations from the mean for each meerkat. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $11$ years $7$ years $49$ years $^2$ $1$ year $-3$ years $9$ years $^2$ $2$ years $-2$ years $4$ years $^2$ $2$ years $-2$ years $4$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{49} + {9} + {4} + {4}} {{4}} $ $ {\sigma^2} = \dfrac{{66}}{{4}} = {16.5\text{ years}^2} $ The average meerkat at the zoo is 4 years old. The population variance is 16.5 years $^2$.